Question

What is the time taken by a particle executing SHM with a time period \( T \) sec from a positive extreme position to half of the amplitude?

• A. \( \dfrac{2T}{12} \, \text{sec} \)
• B. \( \dfrac{T}{12} \, \text{sec} \)
• C. \( \dfrac{6T}{12} \, \text{sec} \)
• D. \( \dfrac{3T}{12} \, \text{sec} \)

Hint:

Use \( x = A \cos(\omega t) \) for SHM starting from the extreme position.

Answer 1

The Correct Option is A

In SHM, displacement is given by: \[ x = A \cos(\omega t) \] Given, \( x = \frac{A}{2} \Rightarrow \frac{A}{2} = A \cos(\omega t) \Rightarrow \cos(\omega t) = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{3} \] Now, \[ \omega = \frac{2\pi}{T} \Rightarrow t = \frac{\pi}{3} \cdot \frac{T}{2\pi} = \frac{T}{6} = \frac{2T}{12} \]